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php - 一个mysql查询的结果在另一个使用php的查询中

转载 作者:行者123 更新时间:2023-11-30 01:29:00 25 4
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我有一个 mysql 查询连接两个表,使用 mysql_fetch_array 将结果放入关联数组中,并使用 while 循环显示“城市名称”。在同一个查询中,我需要使用关联数组的每个相应的“城市名称”再执行一次查询,并显示每个城市的相应标签。

$result = mysql_query("select city.city_ID, city.city_name, city.city_page, city_images.cimage_path from       city, city_images where 

city.city_ID=city_images.city_ID");
while ($row = mysql_fetch_array($result)) {
echo '<div class="Box">';
echo '<a href="' . $row['city_page'] . '">';
echo '<img src=" ' . $row['cimage_path'] . ' " />';
echo '</a>';
echo '<div class="imagetext">';
echo '<a href="' . $row['city_page'] . '">' . $row['city_name'] . '</a>';
echo '</div>';
echo '<div class="Inbox">';
$r = mysql_query("select Tag_Name from tag, city_tag where city_tag.city_ID =$row ['city_ID'] and city_tag.Tag_ID=tag.Tag_ID ");
while ($raw = mysql_fetch_array($r)) {
echo '<div class="boxtags">';
echo '</div>';
}
echo '</div>';
echo '</div>';
}

我需要与当前城市相对应的标签..请帮助

最佳答案

您只需要一个查询...

SELECT c.city_ID
, c.city_name
, c.city_page
, ci.cimage_path
, t.Tag_Name
FROM city c
JOIN city_images ci
ON ci.city_ID = c.city_ID
JOIN city_tag ct
ON ct.city_ID = c.city_ID
JOIN tag t
ON tag.Tag_ID = ct.Tag_ID;

关于php - 一个mysql查询的结果在另一个使用php的查询中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17681612/

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